As I've been reading over some papers on linear functionals, I've noted a large distinction is made between linear operators $T: \mathbb{C}_n[z] \to \mathbb{C}[z]$ and linear operators $T: \mathbb{C}[z] \to \mathbb{C}[z]$, where $\mathbb{C}_n[z] \subset \mathbb{C}[z]$ is the set of complex polynomials of degree of most $n$.
As I've thought about this, it seems to me that if $T: \mathbb{C}[z] \to \mathbb{C}[z]$ then surely we can view $T$ as a mapping from $\mathbb{C}_n[z]$ to $\mathbb{C}[z]$ by considering $T|_{\mathbb{C}_n[z]}$, i.e. the restriction of $T$ to $\mathbb{C}[z]$. However, I can't think of any concrete examples that show the converse is false, i.e. that a linear functional on $\mathbb{C}_n[z]$ cannot be extended to a linear functional on $\mathbb{C}[z]$.
The only examples I can think of do have an extension; for example, if we let $T(p(z)) = \sqrt{p^{(n+1)}(z)}$ then surely $T$ is linear on $\mathbb{C}_n[z]$ and is not linear on $\mathbb{C}[z]$, but $T$ can be extended to $\mathbb{C}[z]$ by defining $T(p(z)) = 1$. My question is thus as follows:
Does there exist a linear functional mapping $\mathbb{C}_n[z]$ to $\mathbb{C}[z]$ that does not extend to a linear functional mapping $\mathbb{C}[z]$ to $\mathbb{C}[z]$?
No, there is not such a linear functional. Because if $\alpha\colon\mathbb{C}_n[z]\longrightarrow\mathbb{C}[z]$ is a linear functional, you can define $A\colon\mathbb{C}[z]\longrightarrow\mathbb{C}[z]$ by$$A(a_0+a_1z+\cdots+a_mz^m)=\begin{cases}\alpha(a_0+a_1z+\cdots+a_mz^m)&\text{ if }m\geqslant n\\\alpha(a_0+a_1z+\cdots+a_nz^n)&\text{ otherwise.}\end{cases}$$Then $A$ is linear and $A|_{\mathbb{C}_n[z]}=\alpha$.
More generally, if $V$ and $W$ are vector spaces and $U$ is a vector subspace of $V$, then every linear map $f\colon U\longrightarrow W$ can be extended to a linear map $F\colon V\longrightarrow W$.